In random sample of 200 chips from a large consignment, 29 items were found to be defective, find the 95% confidence limits for the percentage of defective chips in the consignments.

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Suresh Chand · Accepted Answer

Solution: Sample size (n) = 200 Defective size(d) = 29 The confidence interval is (left ( p pm z_{alpha / 2} sqrt{ frac{pq}{n} } right )) Here, p = (frac{29}{200}) = 0.145 q = 1 - p = 1 - 0.145 = 0.855 For 95% confidence level, the error (α) = 5%, so (1 - frac{α}{2}) = 0.975 and Z0.975 = 1.96 So confidence interval is C.I. = (0.145 pm 1.96 times frac{0.145 times 0.855}{200}) = 0.145 ± 0.0487 = (0.058, 0.1937)

In random sample of 200 chips from a large consignment, 29 items were found to be defective, find the 95% confidence limits for the percentage of defective chips in the consignments.

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In random sample of 200 chips from a large consignment, 29 items were found to be defective, find the 95% confidence limits for the percentage of defective chips in the consignments.

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